 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/39a/b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

r = c

, the traction

σ n

exerted by the outer fluid on the inner fluid is

σ n = - \frac{3 μ B}{c ^{3}} (Ω \times n) .

The inner fluid exerts the opposite traction on the outer fluid. Its couple is therefore

G = - \int_{r = c} x \times (σ n) d S = 3 μ B c^{- 2} \int_{r = c} n \times (Ω \times n) d S .

Using

n \times (Ω \times n) = Ω - n (n \cdot Ω)

and the supplied spherical integral,

\int_{r = c} n_{i} n_{j} d S = \frac{4 π c ^{2}}{3} δ_{ij},

we obtain

\int_{r = c} n \times (Ω \times n) d S = \frac{8 π c ^{2}}{3} Ω .

Thus the radius cancels:

G = 8 π μ B Ω = \frac{8 π μ a ^{3} b ^{3}}{b ^{3} - a ^{3}} Ω .

When

a ≪ b

G \sim 8 π μ a^{3} Ω,

the rotational drag torque for a sphere in an effectively unbounded fluid. If

h = b - a ≪ a

, then

b^{3} - a^{3} \sim 3 a^{2} h, a^{3} b^{3} \sim a^{6},

G \sim \frac{8 π μ a ^{4}}{3 h} Ω,

the inverse-gap growth expected from a thin Couette flow.

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